/*
 *URL: http://codeforces.com/contest/595/problem/B
 *B. Pasha and Phone
 time limit per test1 second
 memory limit per test256 megabytes
 inputstandard input
 outputstandard output
 Pasha has recently bought a new phone jPager and started adding his friends' phone numbers there. Each phone number consists of exactly n digits.

 Also Pasha has a number k and two sequences of length n / k (n is divisible by k) a1, a2, ..., an / k and b1, b2, ..., bn / k. Let's split the phone number into blocks of length k. The first block will be formed by digits from the phone number that are on positions 1, 2,..., k, the second block will be formed by digits from the phone number that are on positions k + 1, k + 2, ..., 2·k and so on. Pasha considers a phone number good, if the i-th block doesn't start from the digit bi and is divisible by ai if represented as an integer.

 To represent the block of length k as an integer, let's write it out as a sequence c1, c2,...,ck. Then the integer is calculated as the result of the expression c1·10k - 1 + c2·10k - 2 + ... + ck.

 Pasha asks you to calculate the number of good phone numbers of length n, for the given k, ai and bi. As this number can be too big, print it modulo 109 + 7.

 Input
 The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ min(n, 9)) — the length of all phone numbers and the length of each block, respectively. It is guaranteed that n is divisible by k.

 The second line of the input contains n / k space-separated positive integers — sequence a1, a2, ..., an / k (1 ≤ ai < 10k).

 The third line of the input contains n / k space-separated positive integers — sequence b1, b2, ..., bn / k (0 ≤ bi ≤ 9).

 Output
 Print a single integer — the number of good phone numbers of length n modulo 109 + 7.
 *
 */

// n : Each phone number consists of exactly n digits
// k : Split the phone numbers into blocks of length k


#include <iostream>
#include <vector>
using namespace std;

int main(){
	int n;
	int k;
	cin >> n >> k;
	int m = n / k;
	vector<long long> aArr(m);
	for(int i = 0; i < m; i++){
		cin >> aArr[i];
	}
	vector<long long> bArr(m);
	for(int i = 0; i < m; i++){
		cin >> bArr[i];
	}

	/* long long maxNum = pow(10, k) - 1; */
	long long ten = 1;
	for(int i = 0; i < k-1; i++)
		ten *= 10;
	long long power = 1;
	for(int i = 0;i < m; i++){
		//firstDigit is the numbers which ok when firstDigit is index
		vector<long long> firstDigit(10,0);
		int a = aArr[i];
		int b = bArr[i];
		long long sum = 0;
		for(int j = 0; j < 10; j++){
			firstDigit[j]  = (ten * (j + 1) - 1) / a + 1;
			if(b != j){
				sum += (firstDigit[j] - (j == 0 ? 0 : firstDigit[j-1]));
			}
		}
		power *= sum;
		power = power % 1000000007;
	}
	cout << power << endl;
	return 0;
}
